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 July 28th, 2017, 02:38 AM #1 Newbie   Joined: Jul 2017 From: Iraq Posts: 18 Thanks: 0 Fourier Transform Hello everyone, am trying to solve this Fourier Trans. problem, here is the original solution >> https://i.imgur.com/eJJ5FLF.png Q/ How did he come up with this result and where is my mistake? All equation are in the above attached picture here is my attempt, part 1>> https://i.imgur.com/DT2tJ0y.jpg part 2>> https://i.imgur.com/jopEoQd.jpg part 3>> https://i.imgur.com/cKXoekT.jpg
 July 28th, 2017, 03:48 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,749 Thanks: 613 Math Focus: Yet to find out. You are out by a factor of $\omega$ in the final line, and perhaps consequently, a sign change. I suspect it was introduced in part 2 somewhere. Thanks from aows61
July 28th, 2017, 04:50 AM   #3
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second attempt

Quote:
 Originally Posted by Joppy You are out by a factor of $\omega$ in the final line, and perhaps consequently, a sign change. I suspect it was introduced in part 2 somewhere.
thanks and here is my second attempt
but also there is a little difference at the final results

https://i.imgur.com/jkVdK7z.jpg

and here is the original solution
https://i.imgur.com/eJJ5FLF.png

can you spot where is the mistake ??

 July 28th, 2017, 05:22 AM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,749 Thanks: 613 Math Focus: Yet to find out. What did you do on 'line 5'
July 28th, 2017, 05:29 AM   #5
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Quote:
 Originally Posted by Joppy What did you do on 'line 5'
No, I just write that word to remember it (not that important).
In the previous solution there is (negative sign) which I forgot to include, but now I've included it,
and, as you can see, the solution is somehow still different from the original one.
Can you explain why?

Last edited by skipjack; July 28th, 2017 at 06:50 AM.

 July 28th, 2017, 06:46 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 $\displaystyle \int_0^\pi\! e^{-j\omega t}\sin(t)dt = \left[-\frac{e^{-j\omega t}(\cos(t) + j\omega\sin(t))}{1 - \omega^2}\right]_0^\pi = \frac{1 + e^{-j\omega\pi}}{1 - \omega^2}$. In the "original" solution, $\theta$ should be $t$. Thanks from aows61
July 28th, 2017, 07:56 AM   #7
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question

Quote:
 Originally Posted by skipjack $\displaystyle \int_0^\pi\! e^{-j\omega t}\sin(t)dt = \left[-\frac{e^{-j\omega t}(\cos(t) + j\omega\sin(t))}{1 - \omega^2}\right]_0^\pi = \frac{1 + e^{-j\omega\pi}}{1 - \omega^2}$. In the "original" solution, $\theta$ should be $t$.
Thanks indeed, but how did you come up with that?
Can you elaborate more?

And is it correct that he reversed the integrals limit in the original solution?

Last edited by skipjack; July 28th, 2017 at 07:59 AM.

 July 28th, 2017, 08:08 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 For that type of integral, see this article. The limits were swapped to compensate for omitting $j^2$ from the denominators. Thanks from aows61
 July 28th, 2017, 08:11 AM #9 Newbie   Joined: Jul 2017 From: Iraq Posts: 18 Thanks: 0 Thanks indeed for your article link, but can you explain more about the results of the integration? Last edited by skipjack; July 28th, 2017 at 09:13 AM.
 July 28th, 2017, 09:16 AM #10 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 What specific thing are you referring to?

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